Limiting Reagent & Water Volume Trick in 60 Seconds!

 

❓ Concept

Limiting Reagent & Water Volume Trick in 60 Seconds!
How do you quickly find how much water forms in a combustion reaction without writing long stoichiometry tables?
This 60-second hack covers:
✔ Limiting reagent shortcut
✔ Quick mole-ratio conversion
✔ Mass → volume trick for liquid water

---

✍️ Short Explanation

Let’s break the entire idea into three ultra-fast steps, using the popular combustion example:

$$\mathrm{C_4H_{10}} + \frac{13}{2} \mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O}$$

This trick works for any reaction that forms liquid water.

---

🔹 Step 1 — Identify the Limiting Reagent FAST

Instead of computing everything, compare “moles available : moles required”.

Example:

• 174 kg butane → $\frac{174000}{58} \approx 3000$
  

• 320 kg O₂ → $\frac{320000}{32} = 10000$
  

Reaction needs 6.5 mol O₂ per mol C₄H₁₀.

Demand for O₂ to consume all butane:

$$3000\times 6.5=19500\text{ mol}$$

Available O₂ = 10000 mol < 19500 mol → O₂ is limiting.

⏱ Shortcut:

Compare available O₂ with 6.5 × moles of butane; whichever falls short is limiting.

---

🔹 Step 2 — Convert Limiting Reagent → Water Produced

Use the simple ratio:

$$6.5\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{O}}_2⟶5\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2\mathrm{O}$$

So per mole O₂:

$${\mathrm{H}}_2\mathrm{O}=\frac{5}{6.5}=0.769$$

Thus water formed:

$$n_{{\mathrm{H}}_2\mathrm{O}}=10000\times 0.769\approx 7692\text{ mol}$$

⏱ Hack:

Divide moles of O₂ by 1.3 (because 6.5/5 = 1.3) to get moles of water instantly.

$$10000÷1.3=7692\text{ mol}$$

---

🔹 Step 3 — Water Mass → Water Volume (Magic Step!)

Since water is liquid, use density ≈ 1 g/mL, i.e.:

$$\text{1 g water}=\text{1 mL water}$$

Mass of water:

$$m=7692\times 18\approx 138500\text{ g}=138.5\text{ kg}$$

Volume:

$$\text{Volume (L)}\approx \text{Mass (kg)}$$

So water volume ≈ 139 liters.

⏱ Hack:

For liquid water, grams → mL and kilograms → liters with no calculation!

---

🧮 Final Result (from Example)

$$\boxed{{\displaystyle \text{Water formed}\approx 139\text{ L}}}$$

---

✅ Conclusion

✔ Limiting reagent? Check stoichiometric needs in one step.
✔ Water formed? Multiply limiting reagent by a fixed ratio.
✔ Volume? Just read mass in kg = liters (for liquid water).

This turns a 2-page stoichiometry question into a 60-second solution.

---